JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 10)
Let f : R $$ \to $$ R be a differentiable function with f(0) = 0. If y = f(x) satisfies the differential equation $${{dy} \over {dx}} = (2 + 5y)(5y - 2)$$, then the value of $$\mathop {\lim }\limits_{n \to - \infty } f(x)$$ is ...........
Answer
0.4
Explanation
We have,
$${{dy} \over {dx}} = (2 + 5y)(5y - 2)$$
$$ \Rightarrow {{dy} \over {25{y^2} - 4}} = dx$$
$$ \Rightarrow {1 \over {25}}\left( {{{dy} \over {{y^2} - {4 \over {25}}}}} \right) = dx$$
On integrating both sides, we get
$${1 \over {25}}\int {{{dy} \over {{y^2} - {{\left( {{2 \over 5}} \right)}^2}}} = \int {dx} } $$
$$ \Rightarrow {1 \over {25}} \times {1 \over {2 \times 2/5}}\log \left| {{{y - 2/5} \over {y + 2/5}}} \right| = x + C$$
$$ \Rightarrow \log \left| {{{5y - 2} \over {5y + 2}}} \right| = 20(x + C)$$
$$ \Rightarrow \left| {{{5y - 2} \over {5y + 2}}} \right| = A{e^{20x}}$$ [$$ \because $$ e20C = A]
when x = 0 $$ \Rightarrow $$ y = 0, then A = 1
$$ \therefore $$ $$\left| {{{5y - 2} \over {5y + 2}}} \right| = {e^{20x}}$$
$$\mathop {\lim }\limits_{x \to - \infty } \left| {{{5f(x) - 2} \over {5f(x) + 2}}} \right| = \mathop {\lim }\limits_{x \to - \infty } {e^{20x}}$$
$$ \Rightarrow \mathop {\lim }\limits_{n \to - \infty } \left| {{{5f(x) - 2} \over {5f(x) + 2}}} \right| = 0$$
$$ \Rightarrow \mathop {\lim }\limits_{n \to - \infty } 5f(x) - 2 = 0$$
$$ \Rightarrow \mathop {\lim }\limits_{n \to - \infty } f(x) = {2 \over 5} = 0.4$$
$${{dy} \over {dx}} = (2 + 5y)(5y - 2)$$
$$ \Rightarrow {{dy} \over {25{y^2} - 4}} = dx$$
$$ \Rightarrow {1 \over {25}}\left( {{{dy} \over {{y^2} - {4 \over {25}}}}} \right) = dx$$
On integrating both sides, we get
$${1 \over {25}}\int {{{dy} \over {{y^2} - {{\left( {{2 \over 5}} \right)}^2}}} = \int {dx} } $$
$$ \Rightarrow {1 \over {25}} \times {1 \over {2 \times 2/5}}\log \left| {{{y - 2/5} \over {y + 2/5}}} \right| = x + C$$
$$ \Rightarrow \log \left| {{{5y - 2} \over {5y + 2}}} \right| = 20(x + C)$$
$$ \Rightarrow \left| {{{5y - 2} \over {5y + 2}}} \right| = A{e^{20x}}$$ [$$ \because $$ e20C = A]
when x = 0 $$ \Rightarrow $$ y = 0, then A = 1
$$ \therefore $$ $$\left| {{{5y - 2} \over {5y + 2}}} \right| = {e^{20x}}$$
$$\mathop {\lim }\limits_{x \to - \infty } \left| {{{5f(x) - 2} \over {5f(x) + 2}}} \right| = \mathop {\lim }\limits_{x \to - \infty } {e^{20x}}$$
$$ \Rightarrow \mathop {\lim }\limits_{n \to - \infty } \left| {{{5f(x) - 2} \over {5f(x) + 2}}} \right| = 0$$
$$ \Rightarrow \mathop {\lim }\limits_{n \to - \infty } 5f(x) - 2 = 0$$
$$ \Rightarrow \mathop {\lim }\limits_{n \to - \infty } f(x) = {2 \over 5} = 0.4$$
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